Question

A 100 kg individual consumes 1200 kcal of food energy a day. Calculate

(a) the altitude change, in m, if the food energy content was converted entirely into lifting the indi-

vidual under normal earth gravity.

(b) the velocity, in m/s, if the food energy content was converted entirely into accelerating the indi-

vidual from rest.

(c) the final temperature, in ◦C, of a 100 kg mass of liquid water initially at the normal human body

temperature and heated with the energy content of the food. You can use a liquid water specific

heat of 4.1 kJ/kg K.

Answer 1

Answer:

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 =  1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C